50. Pow(x, n)

Implement pow(x, n).

 题目：https://leetcode.com/problems/powx-n/description/

Example 1:

Input: 2.00000, 10
Output: 1024.00000

 

Example 2:

Input: 2.10000, 3
Output: 9.26100

求x的n次方

C++https://discuss.leetcode.com/topic/24824/c-4-lines-of-code

class Solution {
public:
    double myPow(double x, int n) {
        if (n==0) return 1;
        if (n==1) return x;
        if (n==-1) return 1/x;
        return myPow(x*x,n/2)*(n%2==0?1:n>0?x:1/x);    //n 是奇数的时候 n/2会少乘上(n - n/2*2 = sign(n)*1)个x。如果n>0，要乘上x,不然就乘上x^-1
    }
};